Answer: The Quotient Square That Becomes a Disk

Key Idea / Intuition

When you pinch the top edge of a square to a single point, you are doing exactly what a cone does: taking a space $A$ (the bottom edge, together with the sides), collapsing one end of $A \times [0,1]$ to a point. The result is a triangle, which is homeomorphic to a disk $D^2$ — or equivalently, to the cone over an interval. The key insight is to write down an explicit homeomorphism and use the closed map lemma: a continuous bijection from a compact space to a Hausdorff space is automatically a homeomorphism.

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Formal Proof / Solution

Step 1: Identify the quotient.

Let $q : [0,1]^2 \to X/{\sim}$ be the quotient map. The equivalence relation collapses $\{(x,1) : x \in [0,1]\}$ to a single point $*$, while all other points are identified only with themselves.

Step 2: Define a candidate homeomorphism.

Consider the closed triangle $T = \{(u, v) \in \mathbb{R}^2 : u \geq 0,\; v \geq 0,\; u + v \leq 1\},$ which is homeomorphic to $D^2$ (it is a compact convex set with nonempty interior).

Define a map $f : [0,1]^2 \to T$ by $f(x, t) = \bigl((1-t)\,x,\; (1-t)(1-x)\bigr).$

Step 3: Check that $f$ is continuous and respects $\sim$.

• $f$ is clearly continuous (polynomial in $x$ and $t$).
• When $t = 1$: $f(x, 1) = (0, 0)$ for all $x$. So all top-edge points map to the apex $(0,0)$ of the triangle — the identification $\sim$ is respected.
• Thus $f$ descends to a continuous map $\bar{f} : X/{\sim} \to T$.

Step 4: Check that $\bar{f}$ is a bijection.

Surjectivity: Given any $(u,v) \in T$, let $s = u+v$. If $s = 0$, then $t = 1$ and any $x$ maps there (this is the apex). If $s > 0$, set $t = 1-s$ and $x = u/s$; then $f(x,t) = (u,v)$.

Injectivity: Suppose $f(x,t) = f(x', t')$, i.e., $(1-t)x = (1-t')x', \qquad (1-t)(1-x) = (1-t')(1-x').$ Adding: $(1-t) = (1-t')$, so $t = t'$. If $t < 1$, then $1-t > 0$ and we get $x = x'$. If $t = t' = 1$, both points are in the collapsed class $[(x,1)]_\sim$, so they represent the same point in $X/\sim$.

Step 5: Apply the closed map lemma.

$X/\sim$ is compact (continuous image of the compact square $[0,1]^2$), and $T$ is Hausdorff (it is a subspace of $\mathbb{R}^2$). A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

Conclusion:

$X/{\sim} \;\cong\; T \;\cong\; D^2.$

Collapsing the top edge of the unit square to a point yields a space homeomorphic to the closed disk (equivalently, a triangle/cone over an interval). $\blacksquare$

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Remark: This is a special case of the general fact that the cone $CX = X \times [0,1] / (X \times \{1\})$ over a contractible compact space $X$ is contractible and, when $X$ is an arc, yields a disk.